find the line(written in standard form) containing the points:?

Question

(-4,7) and (1,9)

Answer 1

First, you need to find the slope.

If you have points (a, b) and (c, d)
the slope is
m = (d - b) / (c - a)

Then you use one of the given points and that slope to write an equation.
m and (a, b)

y - b = m(x - a)

Then manipulate the equation until it's in the form
__x + __y = ____

**********************
(-4, 7) and (1, 9)

m = (9 - 7) / (1 - -4)
= 2 / (1 + 4)
= 2 / 5

Now pick one of the points... I'll pick (1, 9).

y - 9 = (2/5)(x - 1)

multiply both sides by 5 to get rid of fraction
5(y - 9) = 2(x - 1)
5y - 45 = 2x - 2
-2x + 5y - 45 = -2
-2x + 5y = 43
2x - 5y = -43
(Traditionally, the value in front of the x should be positive in standard form.)

Answer 2

The equation of any line passing through the Points (x',y') and (x",y") is :-

y-y' = (y"-y')/(x"-x') *(x-x')

The equation of the line passing through the points (-4,7)and (1,9) is :-

y-7 = (9-7)/1- -4)*(x- -4)

or y-7 = 2/5*(x+4)
or 5y-35=2x+8
or 2x-5y+43=0

What do you mean by standard form. Equation can be written in slope-intercept form or point slope form and the like. This equation can be reduced to any required convenient form.

Answer 3

here x1=-4,y1=7,x2=1,y2=9
so, equation of the line containing points (-4,7) and (1,9)
y-y1= (y2-y1/x2-x1)*(x-x1)
y-7= (9-7/1-(-4) )*(x-(-4) )
y-7= (9-7/(1+4) )*(x+4)
y-7= (2/5)*(x+4)
5(y-7)=2(x+4)
5y-35=2x+8
2x+8-5y+35=0
2x-5y+43=0

Answer 4

change in y: 9 - 7 = 2
change in x: 1 - -4 = 5
slope: 2/5

2x - 5y = c
2(1) - 5(9) = 2 - 45 = -43 .......... or
2(-4) - 5(7) = -8 - 35 = -43

2x - 5y = -43